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Car B is following car A as they are moving along a straight path with vA=40 mph and vB=45 mph. At the moment when the distance between the cars is 45 ft brakes are applied simultaneously in both cars. Car A decelerates with aA=−22 ft/s2 and car B with aB=−20 ft/s2. What is the distance between the cars when they are both stopped?

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hamzaahmeds

Answer:

s = 14.3 ft

Explanation:

First we need to calculate the distances traveled by both the cars. We use third equation of motion for that:

2as = Vf² - Vi²

where,

a = acceleration

s = distance

Vf = Final Velocity

Vi = Initial velocity

FOR CAR A:

Vi = Va = (40 mph)(5280 ft/1 mile)(1 h/3600 s) = 58.66 ft/s

Vf = 0 ft/s

a = aA = - 22 ft/s²

s = sa = ?

Therefore,

2(- 22 ft/s²)(sa) = (58.66 ft/s)² - (0 ft/s)²

sa = 78.2 ft

FOR CAR B:

Vi = Vb = (45 mph)(5280 ft/1 mile)(1 h/3600 s) = 66 ft/s

Vf = 0 ft/s

a = aB = - 20 ft/s²

s = sb = ?

Therefore,

2(- 20 ft/s²)(sb) = (66 ft/s)² - (0 ft/s)²

sb = 108.9 ft

Since, the car A was initially 45 ft ahead of car B. Therefore,

sa = 45 ft + 78.2 ft = 123.2 ft

Now, the distance between the cars will be:

s = sa - sb

s = 123.2 ft - 108.9 ft

s = 14.3 ft

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